Conditions for which $P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z')$

Question

Given two independent draws $(X,Z,Y)$ and $(X',Z',Y')$ from some distribution $P$, I would like to understand under which conditions does the following hold:

$$ P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z'). $$

This clearly holds when $X \perp Y | Z$ for example, but I want to be able to say that the above statement holds if and only if some condition is met.

Answer 1

Let's rewrite the two expressions using conditional expectation $$P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z') \tag{1}$$ \begin{align} & \iff E\left(\mathbb{I}_{ \{ Y \le Y' \}} |X,Z,X',Z'\right) = E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right) \\ & \iff E\left(E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right)|X,X'\right) = E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right) \tag{2} \\ \end{align}

$(2)$ holds true if and only if the variable $E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right)$ is dependant on ($X$ and $X'$). In other words, $E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right)\in \sigma(X,X')$ or there exists a $g$ mesurable such that $E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right) = g(X,X')$ .

Hence, we conclude that the necessary and sufficient condition for $(1)$ is $E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right)$ is dependant on ($X$ and $X'$) or $\exists g:E\left(\mathbb{I}_{ \{ Y \le Y' \}} |Z,Z'\right) = g(X,X')$.

Answer 2

Partial answer / a dependent example:

First of all, lemme simplify by removing $Z, Z'$ from the problem -- i.e. consider examples where $Z$ is independent of everything else.

So, can we have $P(Y \le Y' \mid X, X') = P(Y \le Y')$ even though $X,Y$ are dependent? Here is a classic example, one of the textbook examples illustrating two dependent variables with zero correlation:

• $(X,Y)$ is a random point on the unit circle.

Clearly, $X,Y$ are dependent, and slightly less clearly, they have zero correlation. It is also easy to verify that, for any $X = x, X' = x'$, both probabilities equal $1/2$.

If you want to add back a non-trivial dependence on $Z$, simply pick:

• $(X,Y,Z)$ is a random point on the unit sphere (surface, not interior).

Again, both probabilities equal $1/2$.

Come to think of it, picking from the unit disk (including interior), or from the unit ball (including interior) also satisfies your equation.

Back to your general question, "Conditioned on $Z$, then $X,Y$ have zero correlation" is not sufficient to guarantee your equation, as many examples of zero correlation show. So I still don't know what the general criterion is.